An integral of a rational function on the real line

Question

Show that $ \int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5} \sin \frac{2\pi}{5} $.

How we can show that? Please help me.

Answer 1

If you take the counter-clockwise contour of a closed semi-circle with radius $R$ in the complex plane with a contour along the real line from $-R$ to $+R$, we have

$$I_0=\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx=P.V.\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx$$

$$\lim_{R\to\infty}\oint_{\gamma_R}\frac{1-x}{1-x^5}\ dx=\lim_{R\to\infty}\int_{\text{arc}}\frac{1-x}{1-x^5}\ dx+P.V.\int_{-\infty}^\infty\frac{1-x}{1-x^5}\ dx$$

$$\lim_{R\to\infty}I_{\gamma_R}=\lim_{R\to\infty}I_{\text{arc}}+I_0$$

By Jordan's Lemma, we know that

$$\lim_{R\to\infty}I_{\text{arc}}=0$$

And by Cauchy's residue formula, we know that

$$\lim_{R\to\infty}\oint_{\gamma_r}\frac{1-x}{1-x^5}\ dx=2\pi i\left(\operatorname{Res}(\frac{1-x}{1-x^5},e^{2\pi i/5})+\operatorname{Res}(\frac{1-x}{1-x^5},e^{4\pi i/5})\right)\\=2\pi i\left(-\frac15i\sqrt{\frac12(5+\sqrt5)}\right)=\frac\pi5\sqrt{2(5+\sqrt5)}$$

And $\frac\pi5\sqrt{2(5+\sqrt5)}=\frac{4\pi}5\sin\frac{2\pi}5$, so

$$I_0=\lim_{R\to\infty}I_{\gamma_R}=\frac{4\pi}5\sin\frac{2\pi}5=\frac\pi5\sqrt{2(5+\sqrt5)}$$